The parameter $\sigma^2$ is the (unknown) common variance of the vector components, each of which we assume are normally distributed. For the baseball data we have $45 \cdot Y_i \sim \mathsf{binom}(45,p_i)$, so the normal approximation to the binomial distribution gives (taking $\hat{p_{i}} = Y_{i}$)

$$\hat{p}_{i}\approx \mathsf{norm}(\mathtt{mean}=p_{i},\mathtt{var} = p_{i}(1-p_{i})/45).$$

Obviously in this case the variances are not equal, yet if they had been equal to a common value then we could estimate it with the pooled estimator

$$\hat{\sigma}^2 = \frac{\hat{p}(1 - \hat{p})}{45},$$ where $\hat{p}$ is the grand mean $$\hat{p} = \frac{1}{18\cdot 45}\sum_{i = 1}^{18}45\cdot{Y_{i}}=\overline{Y}.$$ It looks as though this is what Efron and Morris have done (in the 1977 paper).

You can check this with the following R code. Here are the data:

y <- c(0.4, 0.378, 0.356, 0.333, 0.311, 0.311, 0.289, 0.267, 0.244, 0.244, 0.222, 0.222, 0.222, 0.222, 0.222, 0.2, 0.178, 0.156)

and here is the estimate for $\sigma^2$:

s2 <- mean(y)*(1 - mean(y))/45

which is $\hat{\sigma}^2 \approx 0.004332392$. The shrinkage factor in the paper is then

1 - 15*s2/(17*var(y))

which gives $c \approx 0.2123905$. Note that in the second paper they made a transformation to sidestep the variance problem (as @Wolfgang said). Also note in the 1975 paper they used $k - 2$ while in the 1977 paper they used $k - 3$.

I am not quite sure about the $c = 0.212$, but the following article provides a much more detailed description of these data:

Efron, B., & Morris, C. (1975). Data analysis using Stein's estimator and its generalizations. Journal of the American Statistical Association, 70(350), 311-319 (link to pdf)

or more detailed

Efron, B., & Morris, C. (1974). Data analysis using Stein's estimator and its generalizations. R-1394-OEO, The RAND Corporation, March 1974 (link to pdf).

On page 312, you will see that Efron & Morris use an arc-sin transformation of these data, so that the variance of the batting averages is approximately unity:

> dat <- read.table("data.txt", header=T, sep=",")
> yi  <- dat$avg45
> k   <- length(yi)
> yi  <- sqrt(45) * asin(2*yi-1)
> c   <- 1 - (k-3)*1 / sum((yi - mean(yi))^2)
> c
[1] 0.2091971

Then they use c=.209 for the computation of the $z$ values, which we can easily back-transform:

> zi  <- mean(yi) + c * (yi - mean(yi))
> round((sin(zi/sqrt(45)) + 1)/2,3) ### back-transformation
[1] 0.290 0.286 0.282 0.277 0.273 0.273 0.268 0.264 0.259
[10] 0.259 0.254 0.254 0.254 0.254 0.254 0.249 0.244 0.239

So these are the values of the Stein estimator. For Clemente, we get .290, which is quite close to the .294 from the 1977 article.