How can I calculate the R-squared of a simple linear regression only with sample size, the coefficients and its standards deviations? I have the following estimated simple linear regression, which was estimated from a sample of 1217 individuals:
$\hat{y}=1.77663 + 0.0910103x$
where
$\hat{\beta_0} = 1.77663$ and standard deviation of $\hat{\beta_0} = 0.0865446$
$\hat{\beta_1} = 0.0910103$ and standard deviation of $\hat{\beta_1} = 0.0065643$
How can I calculate the R-squared?

The variance (squared standard error) of the slope is
$$\text{Var}(\hat{\beta_1}) = \frac{\hat\sigma^2}{ns_x^2} $$
The explained sum of squares is
$$ESS =  n\hat{\beta_1} ^2  s_x^2$$
The residual sum of squares is
$$RSS = (n-p)\hat\sigma^2$$
where $p$ is the number of parameters that are estimated in the regression.
So we have:
$$\frac{1}{R^2} -1 = \frac{RSS}{ESS} = \frac{n-p}{n} \frac{\hat\sigma^2}{ \hat{\beta_1}^2s_x^2} = \frac{(n-p)\text{Var}(\hat{\beta_1}) }{\hat{\beta_1^2}} = (n-p)\left(\frac{S.E. (\hat\beta_1)}{\hat{\beta_1}} \right)^2$$
And consequently
$R^2 =  \frac{1}{1+ (n-p)\left(\frac{S.E. (\hat\beta_1)}{\hat{\beta_1}} \right)^2}$
### Computational example

### create some data
set.seed(1)
n = 1217
x <- rnorm(n,0,1)
eps = rnorm(n,0,0.1)
y = 1 + 0.1*x + eps

### model
mod <- lm(y~x)
plot(x,y)
lines(x,predict(mod))
out <- summary(mod)

### compare R^2 formula 
1/(1+(n-2)*(out$coefficients[2,2]/out$coefficients[2,1])^2)
out$r.squared

### output
# > 1/(1+(n-2)*(out$coefficients[2,2]/out$coefficients[2,1])^2)
# [1] 0.4852368
# > out$r.squared
# [1] 0.4852368

In your case it is
$$R^2 = \frac{1}{1+(1217-2)\left(\frac{0.0065643}{0.0910103}\right)^2} = 0.1365971$$
