Because the smaller diagonal plane intersection is a square with area of $k^2$, the prism height is $H = k$ and the smaller base diagonal is $d_1 = k$. The base of the prism is a rhomb with area $B = \frac{2}{3}k^2$, from where we obtain
$$\frac{2}{3}k^2 = \frac{d_1 \cdot d_2}{2}$$
or
$$\frac{2}{3}k^2 = \frac{k \cdot d_2}{2}.$$
From the last equality we get that the bigger diagonal of the rhomb is $d_2 = \frac{4}{3}k$. The triangle $\triangle ABO$ is right-angled, so
$$a^2 = \left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2$$
or
$$a^2 = \left(\frac{k}{2}\right)^2 + \left(\frac{4k}{6}\right)^2,$$
from where $a = \frac{5}{6}k$.

a) The volume and area of the prism are
$$V = B \cdot H = \frac{2}{3}k^2 \cdot k = \frac{2}{3}k^3$$
and
$$P = 2B + 4ak = 2 \cdot \frac{2}{3}k^2 + 4 \cdot \frac{5}{6}k^2 = \frac{14}{3}k^2.$$

b) Because the measure numbers of the volume and area of the prism are equal we get
$$\frac{2}{3}k^3 = \frac{14}{3}k^2,$$
hence $k = 7$.