Answer
${\bf 0.173 }\;\rm ms$
Work Step by Step
We have here a $RL$ circuit. The two resistors are in series, so we can reduce the circuit to one resistor of 300-$\Omega$ resistance plus one inductor of 0.075-$\rm H$ that is in series to this equivalent resistor.
Recalling that the decaying current as a function in time is given by
$$I=I_0e^{-t/\tau}$$
where $I_0$ is the maximum current, and $\tau=L/R$, so
$$I=I_0e^{-tR/L}$$
we need to find the time when $I=\frac{1}{2}I_0$;
$$\frac{1}{2} \color{red}{\bf\not} I_0= \color{red}{\bf\not} I_0e^{-tR/L}$$
Hence,
$$\ln\left(\frac{1}{2}\right)=\dfrac{-tR}{L}$$
$$t=\dfrac{L}{-R}\ln\left(\frac{1}{2}\right) $$
Plug the known;
$$t=\dfrac{(0.075)}{-(300)}\ln\left(\frac{1}{2}\right) $$
$$t=\color{red}{\bf 1.73\times 10^{-4}}\;\rm s$$
