Is there any error in my solution? Let $f(x)$ is continuous function for all real values of $x$ and satisfies $$\int^x_0  f(t)dt =\int^1_{x} t^2f(t)dt +\frac{x^{16}}{8}+\frac{x^6}{3}+a$$
Then the value of a is equal to:
(a)  $\frac{-167}{840}$
(b) $\frac{-167}{840}$
(c) $\frac{-167}{840}$
(d) $\frac{-167}{840}$
Is there any error in my following solution:
For $x =1$, $$\int^1_0 f(1)dt =0+\frac{1}{8}+\frac{1}{3}+a = \frac{11}{24}+a$$ 
Differentiating both sides 
$$f(x) = -x^2f(x) +2x^{15}+2x^5$$
$$\implies 2 \int^1_0 \frac{x^{15}+x^5}{1+x^2}dx =\frac{11}{24}+a$$
$$\implies a =-\frac{167}{840}$$

Not quite sure what you are doing, although differentiating to find $f$ is the right way to go:
$$f(x) = -x^2 f(x) + 2 x^{15} + 2 x^5 \implies f(x) = 2 x^5 \frac{1+x^{10}}{1+x^2} = 2 x^5 (1-x^2+x^4-x^6+x^8)$$
Now plug in $x=0$ in the integral equation to find that
$$a = -2 \int_0^1 dx \,(x^7-x^9+x^{11}-x^{13}+x^{15}) = - \left (\frac14-\frac15+\frac16-\frac17+\frac18\ \right ) = -\frac{167}{840}$$
