A) There are $3$ consecutive integers with sum $2015$
$3$ integers are $(x-1),x,(x+1)$
$(x-1)+x+(x+1) = 2015$
$\implies 3x = 2015$
$\implies x = 2015/3$
$=671.66$
$670.66, 671.66, 672.66$ are not integers.
B) There are $4$ consecutive integers with sum $2015$
$4$ integers are $(x+1),(x+2),(x+3),(x+4)$
$(x+1)+(x+2)+(x+3)+(x+4) = 2015$
$\implies 4x +10 = 2015$
$\implies 4x = 2015-10 = 2005$
$\implies x = \dfrac{2005}{4}$
$=501.25$
$502.25, 503.25, 504.25, 505.25$ are not integers.
C) There are $5$ consecutive integers with sum $2015$
$5$ integers are $(x-2),(x-1),x,(x+1),(x+2)$
$ (x-2)+(x-1)+x+(x+1)+(x+2) = 2015$
$\implies 5x = 2015$
$\implies x = 2015/5$
$=403$
$401, 402, 403, 404, 405$ are integers.
The integers are $(403-2),(403-1),403,(403+1),(403+2) \implies 401,402,403,404,405$
Therefore,sum = $401+402+403+404+405 = 2015$
D) There are $3$ consecutive integers with product $2015$
$3$ integers are $(x-1),x,(x+1)$
$(x-1) \times x \times (x+1) = 2015$
$\implies (x^2-x)(x+1) = 2015$
$\implies x^3-x^2+x^2-x = 2015$
$\implies x^3-x = 2015$
$\implies x^3-x-2015 = 0$
it clearly shows that we cannot get integer from this equation.
$\color{green}{∴ \text{Correct option is C) There are 5 consecutive integers with sum 2015}}$