Step 1:

In the orbit of mercury,$e=0.206$ and semi-major axis $a=36$(in millions of miles).

Let $F(ae,0),F'(-ae,0)$ be the foci and $A(a,0),A'(-a,0)$ be the vertices of the elliptical path with the sun at $F$.

Step 2:

The planet is at the greatest distance when it is at $A'$.

The distance $FA'=a(1+e)$

$\qquad\qquad\qquad=36(1+0.206)$

$\qquad\qquad\qquad=36\times 1.206$

$\qquad\qquad\qquad=43.416$ million miles.