smc

Let the coordinates of the quadrilateral be $(n,\ln (n)),(n+1,\ln (n+1)),(n+2,\ln (n+2)),(n+3,\ln (n+3))$. We have by shoelace's theorem, that the area is $$\begin{array}{rl}& \frac{\ln (n)(n+1)+\ln (n+1)(n+2)+\ln (n+2)(n+3)+n\ln (n+3)}{2}-\frac{\ln (n+1)(n)+\ln (n+2)(n+1)+\ln (n+3)(n+2)+\ln (n)(n+3)}{2}\\ & =\frac{\ln \left(\frac{n^{n+1}(n+1{)}^{n+2}(n+2{)}^{n+3}(n+3{)}^n}{(n+1{)}^n(n+2{)}^{n+1}(n+3{)}^{n+2}{n}^{n+3}}\right)}{2}\\ & =\ln \left(\sqrt{\frac{(n+1{)}^2(n+2{)}^2}{n^2(n+3{)}^2}}\right)\\ & =\ln \left(\frac{(n+1)(n+2)}{n(n+3)}\right)\\ & =\ln \left(\frac{91}{90}\right).\end{array}$$ We know that the numerator must have a factor of $13$, so given the answer choices, $n$ is either $12$ or $11$. If $n=11$, the expression $\frac{(n+1)(n+2)}{n(n+3)}$ does not evaluate to $\frac{91}{90}$, but if $n=12$, the expression evaluates to $\frac{91}{90}$. Hence, our answer is $\boxed{12}$.