Since $f(x)$ is given continuous on the closed bounded interval [1,3] $f(x)$ is bounded and assumes all the values lying in the interval [m,M] where
m=min f(x) and M=max f(x)
$1\leq x\leq 3\Rightarrow f(1)\leq f(x)\leq (3)$
If $m\geq M$ then f(x) must assume all the irrational values,we must have m=M i.e f(x) must be constant function.
As f(2)=10 we get
$f(x)=10\forall x\in [1,3]$
$f(1.5)=10$
Hence (a) is the correct answer.