To determine if it is possible to place numbers in the squares of a $3 \times 4 \times 5$ box such that the sum of numbers in every stripe of squares (one square wide) circling the box equals $120$, we need to analyze the sums of the numbers in each stripe.

1. Calculate the total number of squares in each face:
- The $3 \times 4$ face has $3 \times 4 = 12$ squares.
- The $3 \times 5$ face has $3 \times 5 = 15$ squares.
- The $4 \times 5$ face has $4 \times 5 = 20$ squares.

2. Determine the total number of squares in the box:
- The box has 6 faces, so the total number of squares is:
$$2 \times (3 \times 4) + 2 \times (3 \times 5) + 2 \times (4 \times 5) = 2 \times 12 + 2 \times 15 + 2 \times 20 = 24 + 30 + 40 = 94$$

3. Sum of numbers in each stripe:
- Each stripe is a one-square-wide loop around the box. There are three types of stripes:
- Stripes parallel to the $3 \times 4$ faces.
- Stripes parallel to the $3 \times 5$ faces.
- Stripes parallel to the $4 \times 5$ faces.

4. Assign numbers to each face:
- For the $3 \times 4$ faces, place the number $\frac{45}{4}$ in each square.
- For the $3 \times 5$ faces, place the number $3$ in each square.
- For the $4 \times 5$ faces, place the number $\frac{21}{4}$ in each square.

5. Calculate the sum for each type of stripe:
- For the $3 \times 4$ faces:
$$\text{Sum} = 12 \times \frac{45}{4} = 12 \times 11.25 = 135$$
- For the $3 \times 5$ faces:
$$\text{Sum} = 15 \times 3 = 45$$
- For the $4 \times 5$ faces:
$$\text{Sum} = 20 \times \frac{21}{4} = 20 \times 5.25 = 105$$

6. **Check if the sum of numbers in every stripe equals $120$:**
- The sums calculated for the stripes are $135$, $45$, and $105$, none of which equal $120$.

Therefore, it is not possible to place numbers in the squares of the $3 \times 4 \times 5$ box such that the sum of numbers in every stripe of squares circling the box equals $120$.

The final answer is False.