I'm guessing you intended to say [tex]X[/tex] has mean [tex]\mu_X=E[X]=25[/tex] and standard deviation [tex]\sigma_x=\sqrt{\mathrm{Var}[X]}=6[/tex], and [tex]Y[/tex] has means [tex]\mu_Y=E[Y]=30[/tex] and standard deviation [tex]\sigma_Y=\sqrt{\mathrm{Var}[Y]}=4[/tex].
If [tex]W=X+Y[/tex], then [tex]W[/tex] has mean
[tex]E[W]=E[X+Y]=E[X]+E[Y]=55[/tex]
and variance
[tex]\mathrm{Var}[W]=E[(W-E[W])^2]=E[W^2]-E[W]^2[/tex]
Given that [tex]\mathrm{Var}[X]=36[/tex] and [tex]\mathrm{Var}[Y]=16[/tex], we have
[tex]\mathrm{Var}[X]=E[X^2]-E[X]^2\implies E[X^2]=36+25^2=661[/tex]
[tex]\mathrm{Var}[Y]=E[Y^2]-E[Y]^2\implies E[Y^2]=16+30^2=916[/tex]
Then
[tex]E[W^2]=E[(X+Y)^2]=E[X^2]+2E[XY]+E[Y^2][/tex]
[tex]X[/tex] and [tex]Y[/tex] are independent, so [tex]E[XY]=E[X]E[Y][/tex], and
[tex]E[W^2]=E[X^2]+2E[X]E[Y]+E[Y^2]=661+2\cdot25\cdot30+916=3077[/tex]
so that the variance, and hence standard deviation, are
[tex]\mathrm{Var}[W]=3077-55^2=52[/tex]
[tex]\implies\sqrt{\mathrm{Var}[W]}=\sqrt{52}=\boxed{2\sqrt{13}}[/tex]
# # #
Alternatively, if you've already learned about the variance of linear combinations of random variables, that is
[tex]\mathrm{Var}[aX+bY]=a^2\mathrm{Var}[X]+b^2\mathrm{Var}[Y][/tex]
then the variance of [tex]W[/tex] is simply the sum of the variances of [tex]X[/tex] and [tex]Y[/tex], [tex]\mathrm{Var}[W]=36+16=52[/tex], and so the standard deviation is again [tex]\sqrt{52}[/tex].
