For 10 observations $$x_1, x_2, \ldots, x_{10}$$, $$\displaystyle\sum_{i=1}^{10}(x_i + 2)^2 = 180$$ and $$\displaystyle\sum_{i=1}^{10}(x_i - 1)^2 = 90$$. Then their  standard deviation is :

Expand each given square so that the unknowns $$\sum x_i$$ and $$\sum x_i^2$$ appear explicitly.

The first condition is
$$\sum_{i=1}^{10}(x_i+2)^2 =\sum_{i=1}^{10}\bigl(x_i^2+4x_i+4\bigr) =\sum x_i^2+4\sum x_i+40 =180$$ so
$$\sum x_i^2+4\sum x_i=140 \quad\text{−(1)}$$

The second condition is
$$\sum_{i=1}^{10}(x_i-1)^2 =\sum_{i=1}^{10}\bigl(x_i^2-2x_i+1\bigr) =\sum x_i^2-2\sum x_i+10 =90$$ hence
$$\sum x_i^2-2\sum x_i=80 \quad\text{−(2)}$$

Subtract (2) from (1):
$$\bigl(\sum x_i^2+4\sum x_i\bigr) -\bigl(\sum x_i^2-2\sum x_i\bigr) =140-80$$
$$6\sum x_i = 60 \;\Longrightarrow\; \sum x_i = 10$$

Insert $$\sum x_i = 10$$ into (2):
$$\sum x_i^2 - 2(10)=80 \;\Longrightarrow\; \sum x_i^2 = 100$$

The mean is
$$\mu = \frac{\sum x_i}{n} = \frac{10}{10}=1$$

Population variance formula: $$\sigma^2 = \frac{1}{n}\sum(x_i-\mu)^2 = \frac{\sum x_i^2}{n}-\mu^2$$

Therefore
$$\sigma^2 = \frac{100}{10}-1 = 10-1 = 9$$
$$\sigma = \sqrt{9}=3$$

Hence the standard deviation of the ten observations is $$3$$.

Option D which is: $$3$$