Answer
$x=\frac{4}{\sqrt {21}}\approx0.87mi$; see figure and explanations.
Work Step by Step
Step 1. Draw a diagram as shown; the boat is at point $A$ and the village is at $C$, $AB=2mi, BC=6mi$; assume the landing point is at $D$ with BD=$x$ where $x\gt0$.
Step 2. Traveling a distance AD on water $AD=\sqrt {x^2+2^2}$ at a speed of $2mph$ will require a time of $\Delta t_1=\frac{\sqrt {x^2+2^2}}{2}$
Step 3. Traveling a distance DC on land $DC=6-x$ at a speed of $5mph$ will require a time of $\Delta t_2=\frac{6-x}{5}$
Step 4. The total time needed will be $T=\Delta t_1+\Delta t_2=\frac{\sqrt {x^2+4}}{2}+\frac{6-x}{5}$
Step 5. Take its derivative to get $T'=\frac{x}{2\sqrt {x^2+4}}-\frac{1}{5}$. Let $T'=0$, we get $5x=2\sqrt {x^2+4}$, which gives $x=\frac{4}{\sqrt {21}}mi$
Step 6. Checking the signs of $T'$ near the critical point, we have $..(-)..(\frac{4}{\sqrt {21}})..(+)..$ and we confirm the critical point gives a minimum total time.
