A test of two binomial proportions in R, seems appropriate to test $H_0: p_1=p_2$ against $H_a: p_1 \ne p_2.$ The two estimated
proportions are $\hat p_1 = 40/300 = 0.13$ and $\hat p_2 = 200/1000 = 0.20,$ so the observed proportions are different.
Then `prop.test` in R gives a P-value $0.009 < 0.01 = 1\%,$
so the difference is statistically significant at the 1% level.

    prop.test(c(40, 200), c(300,1000), cor=F)

        2-sample test for equality of proportions 
        without continuity correction

    data:  c(40, 200) out of c(300, 1000)
    X-squared = 6.8134, df = 1, p-value = 0.009048
    alternative hypothesis: two.sided
    95 percent confidence interval:
      -0.11243026 -0.02090307
     sample estimates:
       prop 1    prop 2 
    0.1333333 0.2000000 


_Notes:_ (1) Your table is in the correct format for a chi-squared test, shown below. (The different sample sizes are not a problem.) It gives the same P-value as 'prop.test',

    TAB = rbind(c(200,40), c(800, 260))
    TAB
         [,1] [,2]
    [1,]  200   40
    [2,]  800  260

    chisq.test(TAB, cor=F)

            Pearson's Chi-squared test

    data:  TAB
    X-squared = 6.8134, df = 1, p-value = 0.009048

(2) I did not use the continuity correctios for the normal
approximation in either test on account of the sample sizes over 100.

(3) A test similar to `prop.test`, which you can try with hand computation is
described on this [NIST](https://www.itl.nist.gov/div898/software/dataplot/refman1/auxillar/binotest.htm) page.