The situation described in question is shown in figure

The magnetic induction at O, due to the current in large loop is given as

$\begin{array}{l}\text{ where }{B}_0=\frac{{\mu }_{0}i}{2R}\\ \Rightarrow i=\frac{V}{R}=\frac{(4+2.5t)}{\left(2\pi R\right)\rho }\\ \Rightarrow B_0=\frac{{\mu }_0}{2R}\left[\frac{(4+2.5t)}{\left(2\pi R\right)\rho }\right]\end{array}$

Here we consider that the central loop is very small i.e., r << R, so the field may be considered as constant within the loop. Thus within the loop the field is B₀ so the flux through smaller loop is given as

$\varphi =B_0\times \left(\pi r^2\right)=\frac{{\mu }_0}{2R}\left[\frac{(4+2.5t)}{\left(2\pi R\right)\rho }\right]\cdot \pi r^2$

Induced emf in the smaller loop is given as

$e=\left|\frac{d\varphi }{dt}\right|=\frac{{\mu }_0{r}^2}{4R^2\rho }\times \left(2.5\right)$

Thus current i in the smaller loop is given as

$\begin{array}{l}i=\frac{e}{R}=\frac{{\mu }_0{r}^2}{4R^2\rho }\times 2.5\times \left(\frac{1}{2\pi r\rho }\right)\\ \Rightarrow i=\frac{2.5{\mu }_{0}r}{8\pi R^2{\rho }^2}\Rightarrow i=\frac{2.5\times \left(4\pi \times {10}^{-7}\right)\left(0.1\right)}{8\pi \times 1\times {10}^{-8}}=1.25\mathrm{A}\end{array}$