We are given,

The total number of observations, $n = 100$.

Incorrect mean, $\overline x  = 20$.

And, Incorrect standard deviation, $\sigma  = 3$.

So, Mean is given by,

$\overline {\text{X}}  = \frac{1}{{\text{n}}}\sum\limits_{{\text{i}} = 1}^{100} {{{\text{X}}_1}} $

$20 = \frac{1}{{100}}\sum\limits_{{\text{i}} = 1}^{20} {{{\text{X}}_1}} $

$\sum\limits_{i = 1}^{100} {{x_1}}  = 20 \times 100$

$\sum\limits_{i = 1}^{100} {{x_1}}  = 2000$

Thus, the incorrect sum of observations $ = 2000$.

Now, the correct sum of observations $ = 2000 – 21 – 21 – 18$

$ = 2000 – 60$

$ = 1940$

Hence, the correct mean $ = \frac{{correct{\text{ }}sum}}{{100 – 3}}$

$ = \frac{{1940}}{{97}}$

$ = 20$

Also, the standard deviation is given by,

$\sigma  = \sqrt {\frac{1}{{\text{n}}}\sum\limits_{{\text{i}} = 1}^{\text{n}} {{{\text{X}}_1}}  – \frac{1}{{{{\text{n}}^2}}}{{\left( {\sum\limits_{i = 1}^{\text{n}} {\text{X}} } \right)}^2}} $

\[3 = \sqrt {\frac{1}{n}\sum\limits_{i = 1}^n {X_1^2}  – {{(\bar X)}^2}} \]

\[3 = \sqrt {\frac{1}{{100}} \times {\text{ Incorrect }}\sum {{\text{X}}_1^2}  – {{(20)}^2}} \]

\[{\text{Incorrect }}\sum\limits_{i = 1}^n {X_1^2}  = 100\left( {9 + 400} \right)\]

\[{\text{Incorrect }}\sum\limits_{i = 1}^n {X_1^2}  = 40900\]

Now, \[Correct\sum\limits_{{\text{i}} = 1}^{\text{n}} {{\text{X}}_1^2}  = Incorrect\sum\limits_{{\text{i}} = 1}^{\text{n}} {{\text{X}}_1^2}  – {(21)^2} – {(21)^2} – {(18)^2}\]

\[{\text{ = 40900 – 441 – 441 – 324}}\]

\[{\text{ = 40900 – 1206}}\]

\[{\text{ = 39694}}\]

Therefore, the correct standard deviation is,

$ = \sqrt {\frac{{{\text{ Correct }}\sum {{\text{X}}_1^2} }}{{\text{n}}} – {{({\text{ Correct Mean }})}^2}} $

$ = \sqrt {\frac{{39694}}{{97}} – {{(20)}^2}} {\text{ }}$

$ = \sqrt {409.216 – 400} $

$ = 3.036{\text{ }}$