Let the observations be $x_{1}, x_{2}, x_{3}, x _{4}, x_{5} ,$ and $x_{6}$

It is given that mean is $8$ and standard deviation is $4$

Mean, $\bar{x}=\frac{x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}}{6}=8$       …….$(1)$

If each observation is multiplied by $3$ and the resulting observations are $y_{i},$ then

$y_{1}=3 x_{1}$ i.e., $x_{1}=\frac{1}{3} y_{1},$ for $i=1$ to $6$

New Mean, $\bar{y}=\frac{y_{1}+y_{2}+y_{3}+y_{4}+y_{5}+y_{6}}{6}$

$=\frac{3\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}\right)}{6}$

$=3 \times 8$        …….[ Using  $(1)$ ]

$=28$

Standard deviation, $\sigma  = \sqrt {\frac{1}{n}\sum\limits_{i = 1}^6 {{{\left( {{x_1} – \bar x} \right)}^2}} } $

$\therefore {\left( 4 \right)^2} = \frac{1}{6}\sum\limits_{i = 1}^6 {{{\left( {{x_i} – \bar x} \right)}^2}} $

$\sum\limits_{i = 1}^6 {{{\left( {{x_i} – \bar x} \right)}^2}}  = 96$            ……..$(2)$

From $(1)$ and $(2),$ it can be observed that,

$\bar{y}=3 \bar{x}$

$\bar{x}=\frac{1}{3} \bar{y}$

Substituting the values of $x_{1}$ and $\bar{x}$ in $(2),$ we obtain

$\sum\limits_{i = 1}^6 {{{\left( {\frac{1}{3}{y_1} – \frac{1}{3}\bar y} \right)}^2} = 96} $

$ \Rightarrow \sum\limits_{i = 1}^6 {{{\left( {{y_1} – \bar y} \right)}^2} = 864} $

Therefore, variance of new observations $=\left(\frac{1}{6} \times 864\right)=144$

Hence, the standard deviation of new observations is $\sqrt{144}=12$