Find the phase angle of the of the sinusoidal input that will make the natural response zero
My interpretation: Assuming \$i(0) = 0 \space A\$ and
$$ V_s(t) = V_m\sin(\omega t+\beta) $$
It's reasonable to conceive the response \$i(t)\$ as formed by the two components. The Steady state response and the Transient one:
$$ i(t) = K_1\sin(\omega t+\gamma) \space + K_2e^{-\frac{R}{L}t} $$
If there is not a transient reponse, then \$K_2=0\$
In this case
$$ i(0) = K_1\sin(\gamma) = 0 $$
Since \$K_1\$ cannot be zero, then \$\gamma = 0^\circ\$ or \$\gamma = 180^\circ \$
Choosing the first
$$ i(t) = K_1\sin(\omega t) $$
The differential equation representing the circuit is
$$ \frac{d}{dt}i(t) + \frac{R}{L}i(t) = \frac{V_m}{L}\sin(\omega t+\beta) $$
Replacing the expression by \$i(t)\$
$$ K_1\omega \cos(\omega t) + \frac{R}{L}K_1 \sin(\omega t) = \frac{V_m}{L} \sin(\omega t + \beta) $$
Expanding \$\sin(\omega t + \beta)\$ in the right side:
$$ K_1\omega \cos(\omega t) + \frac{R}{L}K_1 \sin(\omega t) = \frac{V_m}{L} \sin(\omega t)\cos(\beta) + \frac{V_m}{L} \sin(\beta)\cos(\omega t) $$
Equating the coefficients in both sides, leads to:
$$
\left\{\begin{matrix}
K_1\omega = \frac{V_m}{L}\sin(\beta) &amp; (1)\\ 
\frac{R}{L}K_1 = \frac{V_m}{L}\cos(\beta) &amp; (2)
\end{matrix}\right.
$$
As \$\tan(\beta) = \frac{\sin(\beta)}{\cos(\beta)}\$, dividing (1) by (2):
$$ \tan(\beta) = \frac{\omega L}{R} $$
So, we get the condition required for the transient response to be null
$$
\begin{equation}
   \boxed{\beta = \arctan(\frac{\omega L}{R})}
\end{equation}
$$
In this case:
$$ \beta \approx 89.24^\circ $$
In other hand, squaring and summing (1) and (2):
$$ K_1^2[\omega^2 + (\frac{R}{L})^2] = (\frac{V_m}{L})^2[\sin^2(\beta) + \cos^2(\beta)]$$
As \$ \sin^2(\beta) + \cos^2(\beta) = 1 \$
$$ K_1 = \frac{V_m}{\sqrt{R^2 + \omega^2 L^2}} $$
In this case:
$$ K_1 \approx 517.63 $$
Finally
$$ i(t) = 517.63\sin(120 \pi t) $$