Answer
No. The expected class grade (or class average grade) after transforming the grades based on the professor's strategy is $\color{blue}{160/3 \approx 53.\overline{3}}.$
Work Step by Step
\begin{align*}
E(g(Y)) &= E(10\sqrt{Y}) \qquad \text{since}\ g(Y) = 10\sqrt{Y} \\
&= \int_{\mathbb{R}} 10\sqrt{y}\cdot f_Y(y)\ dy \qquad [\ \text{Definition of}\ E(10\sqrt{Y})\ ]\\
&= \int_0^{100} 10\sqrt{y}\cdot \left(\frac{1}{5,\!000}(100-y)\right)\ dy \qquad [\ \text{since}\ f_Y(y)=\frac{1}{5,\!000}(100-y),\ y\in[0,100]\ ] \\
&= \frac{10}{5,\!000}\int_0^{100}\left(100y^{1/2}-y^{3/2}\right)\ dy \\
&= \frac{1}{500}\left( \frac{100y^{3/2}}{3/2} - \frac{y^{5/2}}{5/2}\right)_0^{100} \\
&= \frac{1}{500}\left[\left( \frac{200}{3}10^3 - \frac{2}{5}10^5\right) - \left( \frac{200}{3}0^3 - \frac{2}{5}0^5\right)\right] \\
&= \frac{1}{500}\left[2\cdot 10^5\left(\frac{1}{3}-\frac{1}{5}\right) - (0-0)\right] \\
&= \frac{1}{5}\left[2\cdot 10^3\left(\frac{2}{15}\right)\right] \\
&= \frac{1}{5}\left[2\cdot 2^35^3\left(\frac{2}{3\cdot 5}\right)\right] \\
&= \frac{2^5\cdot 5}{3} \\
\color{blue}{E(g(Y))}\ &\color{blue}{= \frac{160}{3} \;=\; 53.\overline{3}}
\end{align*}
The expected value (average) of the tranformed grades of the class using the professor's strategy is $\color{blue}{160/3 \approx 53.\overline{3}}$. The professor's strategy is thus not enough to raise the class average to above $60$.
