Perfect squares

Using the hint :-

one-digit perfect squares $1^2 to 3^2\ = 3$

two-digit perfect squares $4^2 to 9^2 = 2(6) = 12$

three-digit perfect squares $10^2 to 31^2 = 3(22) = 66$

four-digit perfect squares $32^2 to 99^2 = 4(68) = 272$

five-digit perfect squares $100^2 to 316^2 = 5(217) = 1085$

$1438 + 6(x-317+1) = 2014$

$6(x-317+1) = 576$

$x-317+1=96$

$x = 316 + 96$

$x = 412$

The last digit of $(412)^2$ is $2014^{th}$ digit. The $2016^{th}$ digit will be the second digit from the left of $(413)^2.$

$(413)^2 = 170569$

Therefore, $7$ is the $2016^{th} digit.$