Answer
${\bf 53.3}\;\rm mA$
Work Step by Step
$$\color{blue}{\bf [a]}$$
When we close the switch, the current in the $RC$ circuit will flow clockwise through this circuit as the capacitor discharges.
We need to focus on the lower wire of the circuit that is close to the loop.
Let's assume that it is long enough compared to the loop dimensions.
Also, let's assume that the magnetic field due to the other wires of the circuit is negligible.
The direction of the current in this lower wire is from right to left.
The change in the current magnitude due to the discharge of the capacitor will alter the magnetic field exerted by the lower wire on the loop. This change in the magnetic field produces an induced electromotive force (emf) on the loop.
The direction of the magnetic field from the lower wire on the loop is out of the page. This magnetic field decreases with time as the current decreases, so the induced current will oppose this decrease in the magnetic field.
Therefore, according to the right-hand rule, the induced current must be counterclockwise.
Therefore, the direction of the current in the loop is $\bf counterclockwise$.
$$\color{blue}{\bf [b]}$$
We know that the induced current is given by
$$I_{\rm loop}=\dfrac{\varepsilon}{R_{\rm loop}}\tag 1$$
And we know that the induced emf is given by
$$\varepsilon=\left| \dfrac{d\Phi}{dt} \right|\tag 2$$
Recalling, from an example solved in your textbook, that the flux exerted by a current-carrying wire on a loop is given by
$$\Phi=\dfrac{\mu_0Ib}{2\pi}\ln\left[ \dfrac{d+a}{d} \right]$$
where $b$ is the length of the side that faces the lower wire, $a$ is the length of the next side of the loop (width), and $d$ is the distance between the wire and the closest side to it.
Plug into (2),
$$\varepsilon=\left| \dfrac{d }{dt}\dfrac{\mu_0Ib}{2\pi}\ln\left[ \dfrac{d+a}{d} \right] \right| $$
$$\varepsilon=\dfrac{\mu_0b}{2\pi}\ln\left[ \dfrac{d+a}{d} \right] \left| \dfrac{d I}{dt}\right| \tag 3$$
where $I=I_0e^{-t/RC}$ where $I_0$ is the maximum current at the moment we close the switch, $I_0=\dfrac{\Delta V_c}{R}=\dfrac{20}{2}=10\;\rm A$.
Hence,
$$I=10 e^{-t/RC}$$
Plug into (3),
$$\varepsilon=\dfrac{10\mu_0b}{2\pi}\ln\left[ \dfrac{d+a}{d} \right] \left| \dfrac{d e^{-t/RC} }{dt}\right| $$
$$\varepsilon=\dfrac{10\mu_0b}{2\pi}\ln\left[ \dfrac{d+a}{d} \right] \left| \dfrac{-1 }{RC}e^{-t/RC}\right| $$
$$\varepsilon=\dfrac{10\mu_0b}{2\pi}\ln\left[ \dfrac{d+a}{d} \right] \dfrac{ e^{-t/RC} }{RC} $$
Plug into (1),
$$I_{\rm loop}=\dfrac{\dfrac{10\mu_0b}{2\pi}\ln\left[ \dfrac{d+a}{d} \right] \dfrac{ e^{-t/RC} }{RC} }{R_{\rm loop}} $$
Plug the know (at $t=5\;\mu\rm s$):
$$I_{\rm loop}=\dfrac{\dfrac{10(4\pi\times 10^{-7})(0.02)}{2\pi}\ln\left[ \dfrac{0.005+0.01}{0.005} \right] \dfrac{ e^{-(5\mu)/(5\mu)(2)} }{(2)(5\times 10^{-6})} }{0.05} $$
$$I_{\rm loop}=\color{red}{\bf 53.3}\;\rm mA$$
