A random variable following a binomial distribution over [tex]n[/tex] trials with success probability [tex]p[/tex] has PMF
[tex]f_X(x)=\dbinom nxp^x(1-p)^{n-x}[/tex]
Because it's a proper probability distribution, you know that the sum of all the probabilities over the distribution's support must be 1, i.e.
[tex]\displaystyle\sum_xf_X(x)=\sum_{x=0}^n\binom nxp^x(1-p)^{n-x}=1[/tex]
The mean is given by the expected value of the distribution,
[tex]\mathbb E(X)=\displaystyle\sum_xf_X(x)=\sum_{x=0}^nx\binom nxp^x(1-p)^{n-x}[/tex]
[tex]\mathbb E(X)=\displaystyle\sum_{x=1}^nx\frac{n!}{x!(n-x)!}p^x(1-p)^{n-x}[/tex]
[tex]\mathbb E(X)=\displaystyle\sum_{x=1}^n\frac{n!}{(x-1)!(n-x)!}p^x(1-p)^{n-x}[/tex]
[tex]\mathbb E(X)=\displaystyle np\sum_{x=1}^n\frac{(n-1)!}{(x-1)!((n-1)-(x-1))!}p^{x-1}(1-p)^{(n-1)-(x-1)}[/tex]
[tex]\mathbb E(X)=\displaystyle np\sum_{x=0}^n\frac{(n-1)!}{x!((n-1)-x)!}p^x(1-p)^{(n-1)-x}[/tex]
[tex]\mathbb E(X)=\displaystyle np\sum_{x=0}^n\binom{n-1}xp^x(1-p)^{(n-1)-x}[/tex]
[tex]\mathbb E(X)=\displaystyle np\sum_{x=0}^{n-1}\binom{n-1}xp^x(1-p)^{(n-1)-x}[/tex]
The remaining sum has a summand which is the PMF of yet another binomial distribution with [tex]n-1[/tex] trials and the same success probability, so the sum is 1 and you're left with
[tex]\mathbb E(x)=np=126\times0.27=34.02[/tex]
You can similarly derive the variance by computing [tex]\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2[/tex], but I'll leave that as an exercise for you. You would find that [tex]\mathbb V(X)=np(1-p)[/tex], so the variance here would be
[tex]\mathbb V(X)=125\times0.27\times0.73=24.8346[/tex]
The standard deviation is just the square root of the variance, which is
[tex]\sqrt{\mathbb V(X)}=\sqrt{24.3846}\approx4.9834[/tex]
