Correct Answer: $\frac{5}{4}$

Solution :
Since BP and CQ are two medians. So, Q and P are midpoints of AB and AC respectively.
AQ = BQ = $\frac{1}{2}$AB
AP = CP = $\frac{1}{2}$AC
In $\triangle$AQC,
$\angle$A = $90^{\circ}$
By Pythagoras theorem,
⇒ CQ ² = AC ² + QA ²
⇒ 4CQ ² = 4AC ² + 4QA ²
⇒ 4CQ ² = 4AC ² + (2QA) ²
⇒ 4CQ ² = 4AC ² + AB ² ------------(i)
In $\triangle$ BPA,
⇒ BP ² = BA ² + AP ²
⇒ 4BP ² = 4BA ² + 4AP ²
⇒ 4BP ² = 4BA ² + (2AP) ²
⇒ 4BP ² = 4BA ² + AC ² ------------(ii)
Adding (i) and (ii),
⇒ 4CQ ² + 4BP ² = 4AC ² + AB ² + 4BA ² + AC ²
⇒ 4(CQ ² + BP ² ) = 5AC ² + 5BA ² = 5(AC ² + AB ² ) = 5BC ²
⇒ $\frac{BP^2 + CQ^2}{BC^2}$ = $\frac{5}{4}$
Hence, the correct answer is $\frac{5}{4}$.