Thank you Jeremy from Drexel University,
Philadelphia, USA and Andrei, from Tudor Vianu National College,
Bucharest, Romania for two more excellent solutions.
To solve this problem, first I made a
table, and I filled it with the properties of the figures in the
problem.
Stage 
Number of
red triangles

Area
of red
triangles

Number of
white triangles

Area
of white
triangles

$0$ 
$1$ 
$1$ 
$0$ 
$11$ 
$1$ 
$3$ 
$3\over 4$ 
$1$ 
$1{3\over 4}$ 
$2$ 
$3 \times3$ 
${3\over 4}\times {3\over 4}$ 
$1+3$ 
$1\left({3\over 4}\right)^2$ 
From the analysis of the passage from one stage to the next, I made
some observations:
 The number of red triangles at each stage is multiplied by
three to give the number of red triangles at the next stage.
 The total area of the red triangles at each stage is multiplied
by 3/4 to give the total area of the red triangles at the next
stage.
 For each red triangle at stage $n$ one additional white
triangle appears at stage $n+1$. Equivalently the number of white
triangles added at each stage is three times the number of white
triangles added at the previous stage.
 The total area of all the white triangles is [1  (total area
of red triangles)].
In the table below, I summarised the results obtained from
these observations, and I have also calculated the limits for $n
\to \infty$.[Jeremy gave a proof by induction for each of these
formulae]
Stage 
Number
ofred triangles

Area
of red
triangles

Number
of white
triangles

Area
of white
triangles

$n$ 
$3^n$ 
$\left(\frac{3}{4}\right)^n$ 
$1 + 3 +3^2 ... +3^{n1}= {3^n 1\over 2} $ 
$1\left(\frac{3}{4}\right)^n$ 
$n \to \infty$ 
$\infty$ 
$0$ 
$\infty$ 
$1$ 
The total area of the white triangles (that is the triangles
removed) is given by the series: $$\left({1\over 4}\right) +
3\left({1\over 4}\right)^2 + 3^2\left({1\over 4}\right)^3 +...
3^{n1}\left({1\over4}\right)^n = 1  \left({3\over 4}\right)^n$$
and this is as expected from the calculation of the total area of
the red triangles.
In order to calculate the dimension $d$ from the formula $n=m^d$ ,
where $n$ is the number of self similar pieces and $m$ is the
magnification factor, I see that for this problem $n=3$ and $m =
2$. So, $3 = 2^d$ which gives $\log 3 = \log\left(2^d\right)$ and
hence $$d = {\log 3\over \log 2} =\log_2 3\approx 1.58 $$ I see
that the dimension $d$ is between $1$ and $2$, as expected.