The translational energy levels are given by the model of the particle in a three-dimensional box.

$$\epsilon_{tr}=\frac{h^2}{8m}\left(\frac{n_{x}^{2}}{a^2}+\frac{n_{y}^ {2}}{a^2}+\frac{n_{z}^{2}}{a^2}\right)$$ $$q_{tr}=\sum e^{ \beta\epsilon_{tr}}=\sum e^{\frac{\beta h^2}{8m}\left(\frac{n_{x}^{2}}{a^2}+\frac{ n_{y}^{2}}{a^2}+\frac{n_{z}^{2}}{a^2}\right)}=\sum_{n_x =1}^{\infty}e ^{-\frac{\beta h^2n_{x}^{2}}{8ma^2}}\sum_{n_y =1}^{\infty}e^{-\frac{\beta h^2n_{ y}^{2}}{8mb^2}}\sum_{n_z =1}^{\infty}e^{-\frac{\beta h^2n_{z}^{2}}{8mc^2} }$$

Substituting the summations for integrals:

$$q_{tr}=\int_{0}^{\infty}e^{-\frac{\beta h^2n_{x}^{2 }}{8ma^2}}dn_x \int_{0}^{\infty}e^{-\frac{\beta h^2n_{y}^{2}}{8mb^2}}dn_y\int_{0 }^{\infty}e^{-\frac{\beta h^2n_{z}^{2}}{8mc^2}}dn_z$$

Using the integral: $\int_{0}^{ \infty}e^{-\alpha x^2}dx=\frac{1}{2}\left(\frac{\pi}{\alpha}\right)^{1/2}$

$$q_{tr}=\frac{1}{2}\left(\frac{8m\pi}{\beta h^2}\right)^{1/2}a \cdot \frac{1}{2 }\left(\frac{8m\pi}{\beta h^2}\right)^{1/2}b \cdot \frac{1}{2}\left(\frac{8m\pi}{\beta h^2}\right)^{1/2}c$$

Given that $\beta = \frac{1}{k_BT}$ and $V=abc$, the translational partition function is left:

$$q_{tr}=\left(\frac{2\pi mkT}{h^2} \right)^{3/2}V$$