Let, $I_1=$ Inertia of first (rotating) wheel
$\omega_1=$ Angular speed of first wheel
$I_2=3 I_1=$ Inertia of second wheel
$\omega_2=0=$ Angular speed of second wheel
let $\omega=$ Angular speed of coupled wheels.
let $\omega=$ Angular speed of coupled wheels. According to conservation of angular momentum, total initial angular momentum = total final angular momentum
$\begin{aligned} & \Rightarrow I_1 \omega_1+I_2 \times 0=\left(I_1+I_2\right) \omega \\ & \Rightarrow \omega=\frac{I_1 \omega_1}{I_1+I_2}=\frac{I_1 \omega_1}{I_1+3 I_1} \Rightarrow \omega=\frac{\omega_1}{4}\end{aligned}$
Now, initial rotational energy of system is
$K_1=\frac{1}{2} I_1 \omega_1^2+\frac{1}{2} I_2(0)^2$
$\Rightarrow \quad K_1=\frac{1}{2} I_1 \omega_1^2$
Final rotational kinetic energy of system is
$K_2=\frac{1}{2}\left(I_1+I_2\right) \omega^2$
$=\frac{1}{2}\left(I_1+3 I_1\right) \cdot \frac{\omega_1^2}{16}$
$\begin{aligned} & =\frac{1}{2} \times 4 I_1 \times \frac{\omega_1^2}{16} \\ & =\frac{1}{4}\left(\frac{1}{2} I_1 \omega_1^2\right)\end{aligned}$
or $\quad K_2=\frac{1}{4} K_1$
Hence, fraction of lost energy
$\begin{aligned} & =\frac{\Delta K}{K_1}=\frac{K_1-K_2}{K_1}=\frac{\frac{3}{4} K_1}{K_1} \\ & =\frac{3}{4}=0.75\end{aligned}$