Let magnetic field at any time be B. Let upward direction of area vector as positive, the magnetic flux associated with the circuit is

$\begin{array}{l}{\mathrm{\Phi }}_B=\frac{B{L}^2}{2}\text{ and the induced emf is }{\epsilon }_i=-\frac{d{\mathrm{\Phi }}_B}{dt}=-\frac{L^2}{2}\frac{dB}{dt}\text{. }\\ \text{ As, }B=5.0-2.0t\text{, hence, }\frac{dB}{dt}=-2.0\mathrm{T}/\mathrm{s}\\ \text{ It means the induced emf }{\epsilon }_i=-\frac{(1.0{)}^2}{2}(-2.0)=1.0\mathrm{V}\text{. }\end{array}$

As ${\epsilon }_i$ is positive, so the induced emf is  in the same direction as the emf of the battery. The total emf is

${\epsilon }_{\text{total }}=\epsilon +{\epsilon }_i=6.0\mathrm{V}+1.0\mathrm{V}=7.0\mathrm{V}$

The current is in the sense of the total emf (countercloclcvise).
The magnitude of the current in the loop

$i=\frac{{\epsilon }_{\text{total }}}{R}=\frac{7.0}{35.0}=0.2\mathrm{A}$