Since, the pass axes of two polarisers $P_1$ and $P_2$ were kept such that the incident unpolarised beam of intensity $I_0$ gets completely blocked.
This means that both polarisers were kept cross (perpendicular of their pass axes) to each other. When third polariser $P_3$ is introduced between $P_1$ and $P_2$ such that the angle between the pass axes of $P_1$ and $P_3$ is $60^{\circ}$, i.e. $\theta_1=60^{\circ}$.
intensity of polarised light emerging from first polaroid, $I_1=\frac{I_0}{2}$
Intensity of polarised light $\left(I_3\right)$ emerging from polaroid $P_3$ is given by law of Malus, i.e.,
$$
\begin{aligned}
I_3 & =I_1 \cos ^2 \theta, \\
& =\frac{I_0}{2} \cos ^2 60=\frac{I_0}{2} \frac{1}{4} \\
I_3 & =\frac{I_0}{8}
\end{aligned}
$$

Now, the angle between pass axis of $P_3$ and $P_2$, $\theta_2=90^{\circ}-60^{\circ}=30^{\circ}$
$\therefore$ Intensity of polarised light emerging from lost (second) polaroid $P_2$.
$$
I_2=I_3 \cos ^2 30^{\circ}=\frac{I_0}{8} \cdot \frac{3}{4}=\frac{3 I_0}{32}
$$