Answer
$1.0^\circ$
Work Step by Step
We are told that the index of refraction of violet light in this glass is $2\%$ larger than the index of refraction of red light, So
$$n_{\rm violet }=n_{\rm red}+0.02n_{\rm red}$$
$$n_{\rm violet }=1.02n_{\rm red}\tag 1$$
Now we need to apply Snell's law to find the index of refraction of the violet light.
We know that the violet light comes out from the right side of the prism at an angle of $0^\circ$. This means that it falls perpendicular to this right side.
$$n_{air}\sin\theta_{air}=n_{\rm violet }\sin\theta_{\rm 1,violet}$$
So,
$$n_{\rm violet }=\dfrac{n_{air}\sin\theta_{air}}{\sin\theta_{\rm 1,violet}}$$
From the geometry of the figure below, we can see that $\theta_{\rm 1, violet}=90^\circ-60^\circ=\bf 30^\circ$
Hence,
$$n_{\rm violet }=\dfrac{(1.0)\sin50^\circ}{\sin30^\circ}=\bf 1.532$$
Solving (1) for $n_{\rm red}$, and then plug $n_{\rm violet}$,
$$n_{\rm red}=\dfrac{n_{\rm violet }}{1.02}= \dfrac{1.532}{1.02}=\bf 1.502$$
Now we need to work for red,
$$n_{air}\sin\theta_{air}=n_{\rm red}\sin\theta_{\rm 1,red}$$
Hence,
$$\theta_{\rm 1,red}=\sin^{-1}\left[\dfrac{n_{air}\sin\theta_{air}}{n_{\rm red}}\right]$$
Plugging the known;
$$\theta_{\rm 1,red}=\sin^{-1}\left[\dfrac{(1)\sin50^\circ}{1.502}\right]=\bf 30.66^\circ$$
Now we need to find the incident angle of red light on the right side of the prism $\theta_{\rm 2, red}$
From the geometry of the figure below, we can see that
$$\theta_{2,\rm red}=\theta_{1,\rm red}-\theta_{1,\rm violet}$$
$$\theta_{2,\rm red}=30.66^\circ-30^\circ=\bf 0.66^\circ$$
To find $\theta_{3,\rm red}$, we need to apply Snell's law again.
$$\theta_{\rm 3,red}=\sin^{-1}\left[\dfrac{n_{\rm red}\sin\theta_{2,\;\rm red}}{n_{\rm air}}\right]$$
$$\theta_{\rm 3,red}=\sin^{-1}\left[\dfrac{1.502\sin0.66^\circ}{1}\right]=\bf 0.9987^\circ$$
It is obvious that the angle $\theta_{3,\rm red}$ is equal to the angle between red and violet rays.
Thus,
$$\phi\approx \color{red}{\bf 1.0}^\circ $$
