1. Use Faraday’s law

The induced emf in the loop is $|\mathcal E|=\left|\frac{d\phi_B}{dt}\right|$

Given magnetic flux: $\phi_B(t)=10t^2+20t$ with $\phi_B$ in milliweber and $t$ in seconds.

2. Differentiate the flux

$\frac{d\phi_B}{dt}=20t+20$

At $t=5\,\text{s}$, $\frac{d\phi_B}{dt}=20(5)+20=120$

So, $|\mathcal E|=120\,\text{mWb/s}$

Since $1\,\text{Wb/s}=1\,\text{V}$, $120\,\text{mWb/s}=120\,\text{mV}=0.12\,\text{V}$

3. Find the current through the resistor

Given $R=2\,\Omega$, $I=\frac{\mathcal E}{R} = \frac{0.12}{2}=0.06\,\text{A}$

4. Convert into mA

$0.06\,\text{A}=60\,\text{mA}$

5. Final answer

The magnitude of current through the $2\,\Omega$ resistor at $t=5\,\text{s}$ is $\boxed{60\,\text{mA}}$