Given, $T=300 \mathrm{K}, P=2.5\times {10}^3 \mathrm{Pa}$, $V_i=1 {\mathrm{m}}^3 \& V_{\mathit{f}}=3 {\mathrm{m}}^3$, now heat added at constant pressure in terms of the molar specific heat capacity at constant pressure, 

$\mathrm{\Delta }Q=n{C}_p\left(T_2-T_1\right)$

$$\begin{gathered} =n\frac{\gamma R}{\gamma -1}\left(T_2-T_1\right) \\ =n\frac{\gamma }{\gamma -1}\left(R{T}_2-R{T}_1\right) \\ =\frac{\gamma }{\gamma -1}\left(V_2-V_1\right)P \end{gathered}$$

This is a monoatomic gas so the ratio of molar specific heat at constant pressure and constant volume, $\gamma =\frac{C_P}{C_V}=\frac{5}{3}$

$∆Q=\frac{{\displaystyle \frac{5}{3}}}{\left(\frac{5}{3}\right)-1}\times 2.5\times {10}^3\times \left(3-1\right)=12500 \mathrm{J}$

Now using idea gas equation at constant pressure,

$$\begin{gathered} \frac{V_1}{V_2}=\frac{T_1}{T_2} \\ \Rightarrow \frac{1}{3}=\frac{{\displaystyle 300}}{{\displaystyle T_2}} \\ \Rightarrow T_2=900 \mathrm{K} \end{gathered}$$