The top face of the cube is a square, which we label $ABCD$, and we call its centre $O$. Since the cube has edge length 4, then the side length of square $ABCD$ is 4. This means that $O$ is a perpendicular distance of 2 from each of the sides of square $ABCD$, and thus is a distance of $\sqrt{2^{2}+2^{2}}=\sqrt{8}$ from each of the vertices of $ABCD$. These vertices are the farthest points on $ABCD$ from $O$. Since $\sqrt{8} \approx 2.8$, then the loose end of the rope of length 5 can reach every point on $ABCD$, which has area 16. Next, the rope cannot reach to the bottom face of the cube because the shortest distance along the surface of the cube from $O$ to the bottom face is 6 and the rope has length 5. We will confirm this in another way shortly. Also, since the rope is anchored to the centre of the top face and all of the faces are square, the rope can reach the same area on each of the four side faces. Suppose that the area of one of the side faces that can be reached is $a$. Since the rope can reach the entire area of the top face, then the total area that can be reached is $16+4a$. We thus need to determine the value of $a$. Suppose that one of the side faces is square $ABEF$, which has side length 4. Consider the figure created by square $ABCD$ and square $ABEF$ together. We can think of this as an 'unfolding' of part of the cube. When the rope is stretched tight, its loose end traces across square $ABEF$ an arc of a circle centred at $O$ and with radius 5. Notice that the farthest that the rope can reach down square $ABEF$ is a distance of 3, since its anchor is a distance of 2 from $AB$. This confirms that the rope cannot reach the bottom face of the cube since it would have to cross $FE$ to do so. Suppose that this arc cuts $AF$ at $P$ and cuts $BE$ at $Q$. We want to determine the area of square $ABEF$ above $\operatorname{arc} PQ$ (the shaded area); the area of this region is $a$. We will calculate the value of $a$ by determining the area of rectangle $ABQP$ and adding the area of the region between the circular arc and line segment $PQ$. We will calculate this latter area by determining the area of sector $OPQ$ and subtracting the area of $\triangle OPQ$. We note that $PQ=4$. Let $M$ be the midpoint of $PQ$; thus $PM=MQ=2$. Since $\triangle OPQ$ is isosceles with $OP=OQ=5$, then $OM$ is perpendicular to $PQ$. By the Pythagorean Theorem, $OM=\sqrt{OP^{2}-PM^{2}}=\sqrt{5^{2}-2^{2}}=\sqrt{21}$. Thus, the area of $\triangle OPQ$ is $\frac{1}{2} \cdot PQ \cdot OM=\frac{1}{2} \cdot 4 \cdot \sqrt{21}=2 \sqrt{21}$. Furthermore, since $O$ is a distance of 2 from $AB$ and $OM=\sqrt{21}$, then the height of rectangle $ABQP$ is $\sqrt{21}-2$. Thus, the area of rectangle $ABQP$ is $4 \cdot(\sqrt{21}-2)=4 \sqrt{21}-8$. To find the area of sector $OPQ$, we note that the area of a circle with radius 5 is $\pi \cdot 5^{2}$, and so the area of the sector is $\frac{\angle POQ}{360^{\circ}} \cdot 25 \pi$. Now, $\angle POQ=2 \angle POM=2 \sin^{-1}(2/5)$, since $\triangle POM$ is right-angled at $M$ which means that $\sin(\angle POM)=\frac{PM}{OP}$. Thus, the area of the sector is $\frac{2 \sin^{-1}(2/5)}{360^{\circ}} \cdot 25 \pi$. Putting this all together, we obtain $100A=100(16+4a)=1600+400a=1600+400((4\sqrt{21}-8)+\frac{2\sin^{-1}(2/5)}{360^{\circ}} \cdot 25\pi-2\sqrt{21})=1600+400(2\sqrt{21}-8+\frac{2\sin^{-1}(2/5)}{360^{\circ}} \cdot 25\pi)=800\sqrt{21}-1600+\frac{800\sin^{-1}(2/5) \cdot 25\pi}{360^{\circ}} \approx 6181.229$. Therefore, the integer closest to $100A$ is 6181, whose rightmost two digits are 81.