Before I state the question, I want to quickly point out that the question is stated using a constant $\sigma$, while also maintaining that the inner pdf $f$ is a mean 0 and variance 1 density. It's a little confusing, but I believe that the $\sigma$ may refer to some constant and not the standard deviation of $f$.
Suppose the $f$ be a mean 0 density having variance 1. What is the variance associated with the density $g(x) = f(x/\sigma)/\sigma$?
My progress so far is as follows:
$Var(g(x)) = E[(f(x/\sigma)/\sigma)^{2}] - E[f(x/\sigma)/\sigma]^{2}$
I believe the expected value of $g$ is $E[g(x)] = \int_{-\infty}^{\infty}xf(x/\sigma)/\sigma dx = 0 $, and by extension that $E[X]^{2} = 0$.
But I am having difficulty resolving $E[X^{2}]$.
So far I've manipulated it to resemble $\int_{-\infty}^{\infty}x^{2}f(x/\sigma)/\sigma dx$, and I believe I can remove $1/\sigma$ outside the integral to get $(1/\sigma) \int_{-\infty}^{\infty}x^{2}f(x/\sigma)dx$, but I'm uncertain how to proceed from this point.
I suspect the integration should resolve to 1, based on the fact that the question was presented to me as a multiple choice problem with $1/\sigma$ being one of the choices, but I don't know how to prove that, assuming that choice is correct.
The terminology you're using is confusing; I assume that there is a random variable $X$, which has a density $f_X$. The mean of $X$ is $\operatorname{E}[X] = 0$ and the variance of $X$ is $\operatorname{Var}[X] = 1$.
The question, as I have interpreted it, is: if there is a random variable $Y$ with density $f_Y$, such that $f_Y(x) = f_X(x/\sigma)/\sigma$ for $\sigma > 0$, then what is the mean and variance of $Y$?
The answer to this question is that $$f_Y(x) = f_X(x/\sigma)/\sigma \tag{1}$$ implies $$Y = \sigma X, \tag{2}$$ hence $$\operatorname{E}[Y] = \sigma \operatorname{E}[X] = 0, \tag{3}$$ and $$\operatorname{Var}[Y] = \sigma^2 \operatorname{Var}[X] = \sigma^2. \tag{4}$$ We can prove these as follows; for a positive integer $k$,
$$\begin{align} \operatorname{E}[Y^k] &= \int_{y = - \infty}^\infty y^k f_Y(y) \, dy \\ &= \int_{y = -\infty}^\infty y^k f_X(y/\sigma)/\sigma \, dy \\ &= \sigma^{k-1} \int_{y = -\infty}^\infty (y/\sigma)^k f_X(y/\sigma) \, dy \\ &= \sigma^{k-1} \int_{x = -\infty}^\infty x^k f_X(x) \cdot \sigma \, dx & (y = \sigma x, \quad dy = \sigma \, dx) \\ &= \sigma^k \int_{x=-\infty}^\infty x f_X(x) \, dx \\ &= \sigma^k \operatorname{E}[X]. \tag{5} \end{align}$$
Then $(3)$ immediately follows from the case $k = 1$, and $(4)$ follows from setting $k = 2$ and observing $$\operatorname{Var}[Y] = \operatorname{E}[Y^2] - \operatorname{E}[Y]^2 = \sigma^2 \operatorname{E}[X^2] - (\sigma \operatorname{E}[X])^2 = \sigma^2 (\operatorname{E}[X^2] - \operatorname{E}[X]^2) = \sigma^2 \operatorname{Var}[X].$$
Since it seems you're having problems with understanding the above method, I will show step by step how to compute the second moment explicitly:
$$\begin{align} \operatorname{E}[Y^2] &= \int_{y=-\infty}^\infty y^2 f_Y(y) \, dy \\ &= \int_{y=-\infty}^\infty y^2 f_X(y/\sigma)/\sigma \, dy & \text{since by (1), } f_Y(y) = f_X(y/\sigma)/\sigma \\ &= \int_{x=-\infty}^\infty (\sigma x)^2 f_X(\sigma x/\sigma)/\sigma \cdot \sigma \, dx & \text{substitute } y = \sigma x, \quad dy = \sigma \, dx \\ &= \int_{x=-\infty}^\infty \sigma^2 x^2 f_X(x) \, dx \\ &= \sigma^2 \int_{x=-\infty}^\infty x^2 f_X(x) \, dx & \text{factor out the constant} \\ &= \sigma^2 \operatorname{E}[X^2]. \end{align}$$
Since the mean of $X$ is $0$, the variance of $X$ is $$\operatorname{Var}[X] = \operatorname{E}[X^2] - \operatorname{E}[X]^2 = \operatorname{E}[X^2] - 0^2 = \operatorname{E}[X^2];$$ similarly, $\operatorname{Var}[Y] = \operatorname{E}[Y^2]$ since the mean of $Y$ is also $0$ as you already established. Therefore $$\operatorname{Var}[Y] = \sigma^2 \operatorname{Var}[X].$$