You can first use the ideal gas law to calculate $n$:
\begin{align}
pV &= nRT\\
(\pu{101325 Pa})(\pu{0.250 l}) &= n(\pu{8.314 J K-1 mol-1})(\pu{273 K})\\
\end{align}
Solving for "$n$" gives $n = \pu{0.0111577 mol}$.
Now we use the connection between mass $m$, amount of substance $n$, and molar mass mass $M$:
\begin{align}
M &=\frac{m}{n}\\
n &=\frac{m}{M}\\
\pu{0.0111577 mol} &= \frac{\pu{1.78 g}}{M},
\end{align}
which gives a molar mass of $M = \pu{159.53 g mol-1}$.
Looking at the Periodic Table, we can identify the diatomic gas as being bromine gas ($\ce{Br2}$), which has a molar mass of $\pu{159.8 g mol-1}$.