Answer
$5.8\times10^{-2}\;\rm A$
Work Step by Step
We know that the induced current is given by
$$I=\dfrac{\varepsilon}{R}\tag 1$$
And we know that the induced $\varepsilon=\rm emf$ is given by
$$\varepsilon= \dfrac{N_{coil}\Delta \Phi}{\Delta t}= \dfrac{N_{coil}\Delta (BA)}{\Delta t}$$
and we know that the cross-sectional area of the solenoid is constant, so
$$\varepsilon= \dfrac{N_{coil}A_{solenoid}\Delta B_{solenoid}}{\Delta t}\tag 2$$
We also know that the magnetic field of the solenoid is given by
$$B_{solenoid}=\dfrac{\mu_0N_{solenoid} I_{solenoid}}{l_{solenoid}}$$
Thus,
$$\Delta B_{solenoid}=\dfrac{\mu_0N_{solenoid}\Delta I_{solenoid}}{l_{solenoid}}$$
Because $I$ here is the changeable one and all the other variables are constant.
Plugging into (2); we will replace the titles of the coil by $c$ and the solenoid by $s$.
$$\varepsilon= \dfrac{N_{c}A_{s} }{\Delta t} \cdot\dfrac{\mu_0N_{s}\Delta I_{s}}{l_{s}}= \dfrac{\mu_0N_{s}N_{c}A_{s} }{l_{s}\Delta t}\Delta I_{s}$$
Plugging into (1);
$$I= \dfrac{\mu_0N_{c}A_{s} }{R\Delta t}\cdot \dfrac{N_{s}}{l_{s}}
\cdot \Delta I_{s}$$
Plugging the known;
$$I= \dfrac{4\pi \times10^{-7}\cdot 190 \cdot\pi (4.5\times10^{-2} )^2 }{12\cdot 0.10}\cdot \dfrac{230}{10^{-2}}\cdot (2-0)$$
$$I=\color{red}{\bf 5.8\times10^{-2}}\;\rm A$$
And its direction is to the east inside the resistor since the induced current is facing the sudden increase of the field.
