Answer
$\approx 26\;\rm J$
Work Step by Step
We know that the change in the thermal energy of an ideal gas is given by
$$\Delta E_{th}=Q+W$$
So the heat transferred to, or from, the gas is given by
$$Q=\Delta E_{th}-W $$
Recalling that $\Delta E_{th}=nC_{\rm v}\Delta T$, so
$$Q=nC_{\rm v}(T_f-T_i)-W $$
The work magnitude done on the gas is given by the area under the $P-V$ curve, $W=-\int PdV=-\rm Area\; under\; the\; curve$
Hence,
$$Q=nC_{\rm v}(T_f-T_i)-(-\rm Area\; under\; the\; curve) $$
where the Area under the curve is given by the area of a right triangle and a rectangle.
$${\rm Area} =\frac{1}{2}\Delta V\Delta P+P_i\Delta V\\
{\rm Area} =[ \frac{1}{2}(300-100)(4-1)+1(300-100)] \\
=500\times 10^{-6}\times 1.013\times 10^5=\bf 50.65\;\rm J$$
$$Q=nC_{\rm v}(T_f-T_i)+50.65 \tag 1$$
Now we need to find the two temperatures by using the ideal gas law;
$$PV=nRT$$
Thus,
$$T_i=\dfrac{P_iV_i}{nR}=\dfrac{(4\times 1.013\times 10^5)(100\times 10^{-6})}{(0.015)(8.31)}=\bf 325\;\rm K$$
$$T_f=\dfrac{P_fV_f}{nR}=\dfrac{(1\times 1.013\times 10^5)(300\times 10^{-6})}{(0.015)(8.31)}=\bf 244\;\rm K$$
Plugging into (1) and plugging the known;
$$Q=(0.015)(20.4)(244-325)+50.65$$
$$Q=\color{red}{\bf 25.9}\;\rm J$$