The mean and standard deviation of the marks of 10 students were...

Let the observations be ${x}_{1}, {x}_{2}, {x}_{3}, . . . , {x}_{8}, 45, 50$,

Mean$= 50$

$⇒ \dfrac{{x}_{1} + {x}_{2} + {x}_{3} + . . . + {x}_{8} + 45 + 50}{10} = 50$

$⇒ {x}_{1} + {x}_{2} + {x}_{3} + . . . + {x}_{8} = 405 . . . . \left(i\right)$

Hence, new mean

${\left(\overset{¯}{X}\right)}_{new} = \dfrac{405 + 20 + 25}{10} = 45$

Now,

$S . D = \sqrt{\dfrac{\mathop{∑}_{i = 1}^{8}{x}_{i}^{2} + {45}^{2} + {50}^{2}}{10} - {\left(50\right)}^{2}}$

$⇒ 12 = \sqrt{\dfrac{\mathop{∑}_{i = 1}^{8}{x}_{i}^{2} + 4525}{10} - 2500}$

$⇒ \mathop{∑}_{i = 1}^{8}{x}_{i}^{2} = 21915$

Now,

${\left(Variance\right)}_{new} = \dfrac{\mathop{∑}_{i = 1}^{8}{x}_{i}^{2} + {20}^{2} + {25}^{2}}{10} - {\left(45\right)}^{2}$

$⇒ {\left(Variance\right)}_{new} = \dfrac{21915 + {20}^{2} + {25}^{2}}{10} - {\left(45\right)}^{2}$

$⇒ {\left(Variance\right)}_{new} = 2294 - 2025$

$⇒ {\left(Variance\right)}_{new} = 269$

Hence this is the correct answer.