Answer
$n_p\geq1.60$
Work Step by Step
We need to find the index of refraction of the prism $n_p$.
According to Snell's law,
$$\overbrace{n_{air}}^{=1}\;\sin45^\circ=n_p\sin\theta_2$$
Thus,
$$n_p=\dfrac{\sin45^\circ}{\sin\theta_2}\tag 1$$
And to find $n_p$, we need to find $\theta_2$ the angle of refraction, as you see in the figure below.
We need to apply Snell's law to the second surface of the prism.
$$n_p\sin\theta_3=\overbrace{n_{air}}^{=1}\;\;\overbrace{\sin\theta_{4,max}}^{=1}$$
We know that the maximum refracted angle is $90^\circ$ which occurs before the total internally reflected occurs. It is occur when the out light arrow comes out parallel to the right side of our prisme.
Thus,
$$n_p\sin\theta_3\geq 1\tag 2$$
From the geometry of the figure below;
$$\theta_2=90^\circ -\phi_1$$
and
$$\theta_3=90^\circ -\phi_2$$
Adding these two angles together;
$$\theta_2+\theta_3=180^\circ-(\phi_1+\phi_2)\tag 3$$
and from the upper triangle, $65^\circ+\phi_1+\phi_2=180^\circ$
and hence, $\phi_1+\phi_2=180^\circ-65^\circ=\bf 115^{\circ}$
Plugging into (3);
$$\theta_2+\theta_3=180^\circ- 115^\circ=\bf 65^\circ$$
Therefore,
$$\theta_3=65^\circ-\theta_2$$
Plugging into (2);
$$n_p\sin(65^\circ-\theta_2)\geq1 $$
Thus,
$$n_p\geq\dfrac{1}{\sin(65^\circ-\theta_2)}$$
Let's ignore the greater and work for equal only. So plug this result into (1);
$$\dfrac{1}{\sin(65^\circ-\theta_2)}=\dfrac{\sin45^\circ}{\sin\theta_2}$$
Thus,
$$\sin45^\circ \sin(65^\circ-\theta_2)=\sin\theta_2$$
$$\sin45^\circ [\sin 65^\circ\cos\theta_2 -\sin\theta_2\cos65^\circ]=\sin\theta_2$$
$$\sin45^\circ \sin 65^\circ\cos\theta_2 -\sin45^\circ\sin\theta_2\cos65^\circ -\sin\theta_2=0$$
$$\sin45^\circ \sin 65^\circ\cos\theta_2 -\sin\theta_2(\sin45^\circ\cos65^\circ +1)=0$$
$$\sin45^\circ \sin 65^\circ\cos\theta_2 =\sin\theta_2(\sin45^\circ\cos65^\circ +1) $$
$$\dfrac{\sin45^\circ \sin 65^\circ}{ (\sin45^\circ\cos65^\circ +1) }=\dfrac{\sin\theta_2}{\cos\theta_2}=\tan\theta_2$$
Therefore,
$$\theta_2=\tan^{-1}\left[\dfrac{\sin45^\circ \sin 65^\circ }{ (\sin45^\circ\cos65^\circ +1) }\right]=\bf 26.3^\circ$$
Plugging into (1);
$$n_p=\dfrac{\sin45^\circ}{\sin 26.3^\circ} =1.60$$
Therefore, and from all the above, the index of refraction of the prism is greater than or equal to 1.6.
$$n_p\geq\color{red}{\bf 1.60}$$