Since $v_l=v_r$, the robot moves straight ahead.
\[d = vt = 5 \cdot 5 = 25 \cm \]
Note that $\ell=10\cm$ does not come into play here.
Find the angular velocity:
$\omega = \frac{\Delta \theta}{t}
= \frac{\pi}{4} \frac{\rads}{\secs}
$.
Since it's rotation in place, we have $v_l=-v_r$, so $\omega = \frac{v_r-v_l}{\ell} = \frac{2v_r}{10} = \frac{v_r}{5} $ and $\frac{v_r}{5} = \frac{\pi}{4}$. This means that $v_r = \frac{5\pi}{4} \cm/\secs$ and $v_l = -v_r = -\frac{5\pi}{4} \cm/\secs$.
Since it's rotation in place, we have $v_l=-v_r$, so $\omega = \frac{v_r-v_l}{\ell} = \frac{2v_r}{10} = \frac{v_r}{5} $ and $\frac{v_r}{5} = \frac{\pi}{4}$. This means that $v_r = \frac{5\pi}{4} \cm/\secs$ and $v_l = -v_r = -\frac{5\pi}{4} \cm/\secs$.
Find the angular velocity, starting from $\Delta\theta = \omega t$:
\begin{equation*}
\omega
= \frac{\Delta \theta}{t}
= - \frac{\pi/2}{3}
= - \frac{\pi}{6} \frac{\cm}{\secs}
\end{equation*}
Use $\omega$ to find $\ell$:
\begin{equation*}
\ell = \frac{v_r - v_l}{\omega}
= \frac{-3-3}{-\pi/6}
= \frac{36}{\pi} \cm
\end{equation*}