Neyman-Pearson critical constant for variance

Let $X$ be a random sample of size one from a population having $N(0,\sigma^2)$ distribution, where $\sigma > 0$ is an unknown parameter. Let $\Phi(\cdot)$ denote the cumulative distribution function of a standard normal random variable and let $\chi^2_{\nu,\alpha}$ denote the $(1 - \alpha)$-th quantile of the central chi-square distribution with $\nu$ degrees of freedom. It is given that $\Phi(1.96) = 0.975$, $\Phi(1.64) = 0.95$, $\chi^2_{1,0.05} = 3.841$, $\chi^2_{2,0.05} = 5.991$. To test $H_0: \sigma^2 = 1$ against $H_1: \sigma^2 = 2$, using the Neyman-Pearson most powerful test of size 0.05, the critical region is given by $\lambda(X) > c$, where $c \ge 0$ is a constant and $\lambda(x) = \frac{f(x; \sigma^2=2)}{f(x; \sigma^2=1)}$, where $f(x;\sigma^2)$ is the probability density function of a $N(0,\sigma^2)$ distribution. Then the value of $c$ equals ________ (rounded off to two decimal places).